 The Tehran graph is the complete k-partite graph on v vertices, where every partite set has either floor v divided by k or floor v divided by k plus one vertices. It has the greatest possible number of edges for any k-partite graph, but how many edges does it have? Let's start with the easy case and suppose k divides v so that each of the partite sets has v divided by k vertices. There are k choose two pairs of partite sets and each partite set will have v divided by k squared edges between them. So if k divides v, then the number of edges will be and we can do a little bit of algebra and simplify that. So a k-partite set with v vertices can have at most this many edges when k divides v. If k does not divide v, the actual number of edges will be less. We can prove this, but you should do it. But here's a hint. Suppose you add our additional vertices so that v plus r is divisible by k, then there'll be this many edges. What happens when you remove the r extra vertices? So now suppose we have v vertices and n minus one partite graph with v vertices will be k and free. Equivalently, it will have no n-clicks. The Teuron graph is the largest n minus one partite graph on v vertices and so it's n-click free. But it's conceivable that a larger non n minus one partite graph is also n-click free. Now while it is possible that doesn't give us a good starting point. So let's work under the assumption that the Teuron graphs are the largest graphs with no n-clicks. So tkv will have at most this many edges and no n-clicks for any n greater than k. So let's conjecture that any graph with v vertices and at most this many edges will have no n-clicks for n greater than k. We'll prove this using induction. So the theorem is clearly true for v equals one, two, three up to k minus one vertices. So now suppose that all graphs with v equals k vertices and this many edges is Yeah, the problem is we have a lot of variables running around and while we could and should prove this theorem generally, we'll have to choose our variables carefully. So we'll leave that for homework and prove the theorem for a specific value of k, that is the number of parts. Remember concrete doesn't hurt. Since we started by trying to avoid k5 as a subgraph, let's try to avoid five cliques. The corresponding Teuron graph would be t4v and the number of edges must be no more than. So suppose that all graphs with up to k vertices, but this many edges are free of five cliques. So now let's consider a graph with k plus one vertices. So suppose g is maximal in edges, but has no five cliques. Then it must have at least one four-click. That's because we could add one edge to g and get a graph with a five-click. Conversely, we could remove an edge from a five-click to get g, but in that case we'd be left with several four-cliques. Now consider two induced subgraphs of g, a the subgraph corresponding to the four-click and note that a has four vertices and six edges. And we also have g minus a. Now this subgraph g minus a will have k plus one minus four k minus three vertices. While we don't know how many edges it has, we know it doesn't have any five cliques. So by our induction assumption, it has at most this many edges. Now besides the edges in a and g minus a, g will also have edges joining these two subgraphs. But again, since g has no five cliques, then no vertex can connect to all vertices in a. So each of the k minus three vertices in g minus a will have at most three edges to a. So there are at most three k minus three additional edges. Consequently, the number of edges in g is at most the edges in a, plus the edges in g minus a, plus the edges joining a and g minus a. In other words, that's six plus this amount plus this amount. Now we'll take this rather nightmarish-looking expression and we'll do a little algebra, add a little more. Now if we rewrite our terms this way, we have the square of k minus three plus four, which simplifies to, which is exactly what we need if our theorem was going to be true for k plus one. And so our theorem is true for k plus one vertices proving the induction step. And again, we can do the proof generally giving us the following. The k-partite graph on v vertices has at most this many edges and no k plus one clique. It is the largest graph with this many vertices and no k plus one clique. And this is known as Tehran's theorem. The Tehran graph is the largest graph with v vertices and no k plus one clique. Let's verify Tehran's theorem for a graph with, how about 20 vertices and no seven cliques. So our theorem claims we can find a graph with as many as 166 edges, v round down since the number of edges must be a whole number, and no seven cliques. Now we can also construct the Tehran graph T620. And that'll be the complete six-partite graph k 4, 4, 3, 3, 3, 3. And this graph will have 166 edges and no seven cliques. So again, we have a graph with 166 edges and no seven cliques, the Tehran graph, and our theorem says that this is the most we can have. Adding any more edges will give us a seven clique.